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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Find $\sin \large\frac{x}{2},$$\cos \large\frac{x}{2}$ and $\tan \large\frac{x}{2}$ in each of the following $\tan x=-\large\frac{4}{3}$,$x$ in quadrant II

$\begin{array}{1 1} \sin \frac{x}{2} =\frac{2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{-1}{\sqrt 5} \\ \sin \frac{x}{2} =\frac{-2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{1}{\sqrt 5} \\ \sin \frac{x}{2} =\frac{-2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{-1}{\sqrt 5} \\ \sin \frac{x}{2} =\frac{2}{\sqrt 5} ; \cos \frac{x}{2} =\frac{1}{\sqrt 5} \end{array}$

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  • $\cos x=\large\frac{-1}{\sqrt{1\tan^2x}}$
Since $x$ lies in the second quadrant
$\therefore \cos x$ is negative.
$\therefore \cos x=\large\frac{-1}{\sqrt{1+\tan^2x}}=-\frac{1}{\sqrt{1+\Large\frac{16}{9}}}=-\frac{3}{5}$
Now ,$x$ lies in II quadrant
$\Rightarrow \large\frac{\pi}{2}$$ < x < \pi\Rightarrow \large\frac{\pi}{4} $$< \large\frac{x}{2}$$ < \large\frac{\pi}{2}$
$\Rightarrow \large\frac{x}{2}$ lies in first quadrant.
$\Rightarrow \sin \large\frac{x}{2}$$,\cos \large\frac{x}{2}$ and $\tan \large\frac{x}{2}$ are positive
$\therefore\cos\large\frac{x}{2}=\sqrt{\large\frac{1+\cos x}{2}}=\sqrt{\large\frac{1-\Large\frac{3}{5}}{2}}=\sqrt{\large\frac{5-3}{2\times 5}}=\frac{1}{\sqrt 5}$
$\therefore\sin\large\frac{x}{2}=\sqrt{\large\frac{1-\cos x}{2}}=\sqrt{\large\frac{1+\Large\frac{3}{5}}{2}}=\sqrt{\large\frac{5+3}{2\times 5}}=\frac{2}{\sqrt 5}$
$\therefore\tan\large\frac{x}{2}=\sqrt{\large\frac{1-\cos x}{1+\cos x}}=\sqrt{\large\frac{1+\Large\frac{3}{5}}{1-\Large\frac{3}{5}}}=\sqrt{\large\frac{5+3}{5-3}}=$$\sqrt{4}=2$
answered Apr 18, 2014 by sreemathi.v
 

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